A CD has a lot of little marks on it to represent bits, and you can use a laser to read those bits.
A CD burner can use a more powerful laser to make those marks in the CD, as well as read them.
Correct me if I'm wrong so far (as I'm sure I am).
So how does a CD re-writer remove those marks so that they can write a different set of bits?
"Ignorance is bliss, but I tend to get screwed over."
More about :writing work
January 6, 2002 2:04:51 AM
Ok, in the most basic terms the way you think a cd works isn't entirely true, it's what the "marks" do rather than the "marks" themselves. Basically a pressed cd has pits pressed onto it, if the pit is parallel to the reflective material, the laser is reflected back and that pit is a 1, if it has a slant to it the laser gets scattered and that pit is a 0.
A cdr uses a dye that is opaque to infrared light. When it is hit with a light of certain intensity and type, the dye becomes transparent. So if the dye isn't illuminated, it doesn't get reflected by the reflective layer. Once illuminated it let's light through to be reflected.
A cdrw uses an alloy that changes states when hit by strong light. One state is crystalline that refracts light, the other reflects it back to the laser.
"cd-r disks have a metallic layer that can only change phase once"
"hit with a less intense laser than the 'burn' laser "
"after many writes and wipes the layer eventually breaks down "
January 9, 2002 1:33:38 AM
"cd-r disks have a metallic layer that can only change phase once. from crystalline to amorphous."
This isn't true at all, the dye does all the work on a cdr. The dye either filters out the infrared laser so that it doesn't hit the reflective layer or it is hit by the write laser and then lets the read laser through to be reflected.
"course after many writes and wipes the layer eventually breaks down and no longer returns to the crystalline state. at this point the cd-rw wears out."
Actually, each erase doesn't completely change the material back to a crystalline state, some anamorphic material is left behind. Therefore, with each erase the disk comes closer to being equally srystalline and anamorphic. Once this happens the layer still returns to crystalline, just not close enough for the reader to make a determination between a 1 and a 0.